volume 4, issue 5, article 93, 2003.
Received 03 April, 2003;
accepted 21 October, 2003.
Communicated by:S.S. Dragomir
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Journal of Inequalities in Pure and Applied Mathematics
AN ENTROPY POWER INEQUALITY FOR THE BINOMIAL FAMILY
PETER HARREMOËS AND CHRISTOPHE VIGNAT
Department of Mathematics, University of Copenhagen, Universitetsparken 5, 2100 Copenhagen, Denmark.
EMail:moes@math.ku.dk
University of Copenhagen and Université de Marne la Vallée, 77454 Marne la Vallée
Cedex 2, France.
EMail:vignat@univ-mlv.fr
c
2000Victoria University ISSN (electronic): 1443-5756 043-03
An Entropy Power Inequality for the Binomial Family
Peter Harremoës and Christophe Vignat
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Abstract
In this paper, we prove that the classical Entropy Power Inequality, as derived in the continuous case, can be extended to the discrete family of binomial random variables with parameter1/2.
2000 Mathematics Subject Classification:94A17
Key words: Entropy Power Inequality, Discrete random variable
The first author is supported by a post-doc fellowship from the Villum Kann Ras- mussen Foundation and INTAS (project 00-738) and Danish Natural Science Coun- cil.
This work was done during a visit of the second author at Dept. of Math., University of Copenhagen in March 2003.
Contents
1 Introduction. . . 3
2 Superadditivity . . . 4
3 An Information Theoretic Inequality . . . 5
4 Proof of the Main Theorem . . . 9
5 Acknowledgements. . . 11 References
An Entropy Power Inequality for the Binomial Family
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1. Introduction
The continuous Entropy Power Inequality
(1.1) e2h(X)+e2h(Y) ≤e2h(X+Y)
was first stated by Shannon [1] and later proved by Stam [2] and Blachman [3].
Later, several related inequalities for continuous variables were proved in [4], [5] and [6]. There have been several attempts to provide discrete versions of the Entropy Power Inequality: in the case of Bernoulli sources with addition modulo 2, results have been obtained in a series of papers [7], [8], [9] and [11].
In general, inequality (1.1) does not hold when X and Y are discrete ran- dom variables and the differential entropy is replaced by the discrete entropy: a simple counterexample is provided whenX andY are deterministic.
In what follows, Xn ∼ B n,12
denotes a binomial random variable with parametersnand 12,and we prove our main theorem:
Theorem 1.1. The sequenceXnsatisfies the following Entropy Power Inequal- ity
∀m, n≥1, e2H(Xn)+e2H(Xm) ≤e2H(Xn+Xm).
With this aim in mind, we use a characterization of the superadditivity of a function, together with an entropic inequality.
An Entropy Power Inequality for the Binomial Family
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2. Superadditivity
Definition 2.1. A functionnyYnis superadditive if
∀m, n Ym+n≥Ym+Yn.
A sufficient condition for superadditivity is given by the following result.
Proposition 2.1. If Ynn is increasing, thenYnis superadditive.
Proof. Takemandnand supposem≥n. Then by assumption Ym+n
m+n ≥ Ym m or
Ym+n ≥Ym+ n mYm. However, by the hypothesism≥n
Ym m ≥ Yn
n so that
Ym+n≥Ym+Yn.
In order to prove that the function
(2.1) Yn=e2H(Xn)
is superadditive, it suffices then to show that functionn y Ynn is increasing.
An Entropy Power Inequality for the Binomial Family
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3. An Information Theoretic Inequality
Denote asB ∼Ber(1/2)a Bernoulli random variable so that
(3.1) Xn+1 =Xn+B
and
(3.2) PXn+1 =PXn ∗PB = 1
2(PXn+PXn+1), wherePXn ={pnk}denotes the probability law ofXnwith
(3.3) pnk = 2−n
n k
.
A direct application of an equality by Topsøe [12] yields (3.4) H PXn+1
= 1
2H(PXn+1) + 1
2H(PXn) + 1
2D PXn+1||PXn+1 +1
2D PXn||PXn+1 .
Introduce the Jensen-Shannon divergence (3.5) J SD(P, Q) = 1
2D
P
P +Q 2
+1
2D
Q
P +Q 2
and remark that
(3.6) H(PXn) = H(PXn+1),
An Entropy Power Inequality for the Binomial Family
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since each distribution is a shifted version of the other. We conclude thus that
(3.7) H PXn+1
=H(PXn) +J SD(PXn+1, PXn),
showing that the entropy of a binomial law is an increasing function ofn.Now we need the stronger result that Ynn is an increasing sequence, or equivalently that
(3.8) log Yn+1
n+ 1 ≥logYn n or
(3.9) J SD(PXn+1, PXn)≥ 1
2logn+ 1 n .
We use the following expansion of the Jensen-Shannon divergence, due to B.Y.
Ryabko and reported in [13].
Lemma 3.1. The Jensen-Shannon divergence can be expanded as follows J SD(P, Q) = 1
2
∞
X
ν=1
1
2ν(2ν−1)∆ν(P, Q) with
∆ν(P, Q) =
n
X
i=1
|pi −qi|2ν (pi +qi)2ν−1.
This lemma, applied in the particular case whereP =PXn andQ =PXn+1 yields the following result.
An Entropy Power Inequality for the Binomial Family
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Lemma 3.2. The Jensen-Shannon divergence betweenPXn+1 andPXn can be expressed as
J SD(PXn+1, PXn) =
∞
X
ν=1
1
ν(2ν−1)· 22ν−1 (n+ 1)2νm2ν
B
n+ 1,1 2
,
where m2ν B n+ 1,12
denotes the order2ν central moment of a binomial random variableB n+ 1,12
.
Proof. DenoteP =pi, Q=p+i andp¯i = (pi+p+i )/2. For the term∆ν(PXn+1, PXn) we have
∆ν(PXn+1, PXn) =
n
X
i=1
p+i −pi
2ν
p+i +pi2ν−1 = 2
n
X
i=1
p+i −pi
p+i +pi 2ν
¯ pi
and
p+i −pi p+i +pi
= 2−n i−1n
−2−n ni 2−n i−1n
+ 2−n ni = 2i−n−1 n+ 1 so that
∆ν(PXn+1, PXn) = 2
n
X
i=1
2i−n−1 n+ 1
2ν
¯ pi
= 2 2
n+ 1 2ν n
X
i=1
i− n+ 1 2
2ν
¯ pi
= 22ν+1 (n+ 1)2νm2ν
B
n+ 1,1 2
.
An Entropy Power Inequality for the Binomial Family
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Finally, the Jensen-Shannon divergence becomes J SD(PXn+1, PXn) = 1
4
+∞
X
ν=1
1
ν(2ν−1)∆ν(PXn+1, PXn)
=
+∞
X
ν=1
1
ν(2ν−1)· 22ν−1 (n+ 1)2νm2ν
B
n+ 1,1 2
.
An Entropy Power Inequality for the Binomial Family
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4. Proof of the Main Theorem
We are now in a position to show that the function n y Ynn is increasing, or equivalently that inequality (3.9) holds.
Proof. We remark that it suffices to prove the following inequality
(4.1)
3
X
ν=1
1
ν(2ν−1) · 22ν−1 (n+ 1)2νm2ν
B
n+ 1,1 2
≥ 1 2log
1 + 1
n
since the terms ν > 3in the expansion of the Jensen-Shannon divergence are all non-negative. Now an explicit computation of the three first even central moments of a binomial random variable with parametersn+ 1and 12 yields
m2 = n+ 1
4 , m4 = (n+ 1) (3n+ 1)
16 and m6 = (n+ 1) (15n2+ 1)
64 ,
so that inequality (4.1) becomes 1
60
30n4+ 135n3+ 245n2+ 145n+ 37
(n+ 1)5 ≥ 1
2log
1 + 1 n
.
Let us now upper-bound the right hand side as follows log
1 + 1
n
≤ 1 n − 1
2n2 + 1 3n3 so that it suffices to prove that
1
60· 30n4+ 135n3+ 245n2+ 145n+ 37
(n+ 1)5 − 1
2 1
n − 1
2n2 + 1 3n3
≥0.
An Entropy Power Inequality for the Binomial Family
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Rearranging the terms yields the equivalent inequality 1
60· 10n5−55n4−63n3−55n2−35n−10 (n+ 1)5n3 ≥0 which is equivalent to the positivity of polynomial
P (n) = 10n5−55n4−63n3−55n2−35n−10.
Assuming first thatn ≥7,we remark that P(n)≥10n5−n4
55 + 63 6 + 55
62 + 35 63 +10
64
=
10n− 5443 81
n4
whose positivity is ensured as soon asn ≥7.
This result can be extended to the values1≤n ≤6by a direct inspection at the values of functionn y Ynn as given in the following table.
n 1 2 3 4 5 6
e2H(Xn)
n 4 4 4.105 4.173 4.212 4.233 Table 1: Values of the functionny Ynn for1≤n≤6.
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5. Acknowledgements
The authors want to thank Rudolf Ahlswede for useful discussions and pointing our attention to earlier work on the continuous and the discrete Entropy Power Inequalities.
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References
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[9] H.S. WITSENHAUSEN, Entropy inequalities for discrete channels, IEEE Trans. Inform. Theory, IT-20 (1974), 610–616.
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[10] R. AHLSWEDE AND J. KÖRNER, On the connection between the en- tropies of input and output distributions of discrete memoryless channels, Proceedings of the Fifth Conference on Probability Theory, Brasov, Sept.
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